How to evaluate $\int \frac{\log x}{x^2}$?

Question

I want to know how to evaluate $\int \frac{\log x}{x^2}$. Using by parts, and after moving terms, we get something like $$2 \int\frac{\log x}{x^2} = \frac{(\log x)^2}{x} + \int \frac{(\log x)^2}{x^2}$$ Using by parts again gives $$\int \frac{(\log x)^2}{x^2} = \frac{2}{3}(\log x)^3 - \int \frac{(\log x)^3}{x^2}$$ At this point, I give up, since it looks like this is leading to nowhere.

Wxmaxima suggests the answer should be $-\frac{\log x}{x} - \frac{1}{x}$ but I can neither derive it nor find it on the Internet. Thank for your help!

Answer 1

$$\int\frac{\ln x}{x^2}dx$$ $$\int(\ln x) \frac{1}{x^2}dx$$ $$\ln x. (\frac{-1}{x})+\int\frac{1}{x^2}dx$$ $$\frac{-\ln x}{x^2}-\frac{1}{x}+C$$

Answer 2

Well, using integration by parts:

$$\int\frac{\ln\left(x\right)}{x^2}\space\text{d}x=-\frac{\ln\left(x\right)}{x}+\int\frac{1}{x^2}\space\text{d}x\tag1$$

Answer 3

Use $$u=\log(x)\implies u'=\frac{dx}x$$ $$v'=\frac {dx}{x^2}\implies v=-\frac 1x$$